結果

問題 No.2763 Macaron Gift Box
ユーザー 👑 binap
提出日時 2024-04-27 01:07:52
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 190 ms / 3,000 ms
コード長 2,549 bytes
コンパイル時間 5,138 ms
コンパイル使用メモリ 262,736 KB
最終ジャッジ日時 2025-02-21 09:12:50
ジャッジサーバーID
(参考情報)
judge1 / judge4
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ファイルパターン 結果
sample AC * 2
other AC * 15
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ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
using mint = modint998244353;
// Thanks for Luzhiled-san's website
// https://ei1333.github.io/luzhiled/snippets/math/formal-power-series.html
template< typename T >
struct FormalPowerSeries : vector< T > {
using vector< T >::vector;
using P = FormalPowerSeries;
template<class...Args> FormalPowerSeries(Args...args): vector<T>(args...) {}
FormalPowerSeries(initializer_list<T> a): vector<T>(a.begin(),a.end()) {}
using MULT = function< P(P, P) >;
static MULT &get_mult() {
static MULT mult = [&](P a, P b){
P res(convolution(a, b));
return res;
};
return mult;
}
static void set_fft(MULT f) {
get_mult() = f;
}
P operator+(const P &r) const { return P(*this) += r; }
P operator*(const P &r) const { return P(*this) *= r; }
P operator*(const T &v) const { return P(*this) *= v; }
P &operator+=(const P &r) {
if(r.size() > this->size()) this->resize(r.size());
for(int i = 0; i < int(r.size()); i++) (*this)[i] += r[i];
return *this;
}
P &operator*=(const T &v) {
const int n = (int) this->size();
for(int k = 0; k < n; k++) (*this)[k] *= v;
return *this;
}
P &operator*=(const P &r) {
if(this->empty() || r.empty()) {
this->clear();
return *this;
}
assert(get_mult() != nullptr);
return *this = get_mult()(*this, r);
}
P pre(int sz) const {
return P(begin(*this), begin(*this) + min((int) this->size(), sz));
}
// F(0) must not be 0
P inv(int deg = -1) const {
assert(((*this)[0]) != T(0));
const int n = (int) this->size();
if(deg == -1) deg = n;
P ret({T(1) / (*this)[0]});
for(int i = 1; i < deg; i <<= 1) {
ret = (ret * T(2) + ret * ret * pre(i << 1) * T(-1)).pre(i << 1);
}
return ret.pre(deg);
}
};
using fps = FormalPowerSeries<mint>;
int main(){
int n, k;
cin >> n >> k;
assert(1 <= k && k <= n && n <= 200000);
auto pentagonal = [&](int x){
return x * (3 * x - 1) / 2;
};
fps f(n + 1);
{
int i = 0;
mint c = 1;
while(pentagonal(i) <= n){
f[pentagonal(i)] += c;
c *= -1;
i++;
}
}
{
int i = -1;
mint c = -1;
while(pentagonal(i) <= n){
f[pentagonal(i)] += c;
c *= -1;
i--;
}
}
fps g(n + 1);
for(int i = 0; i * (k + 1) <= n; i++) g[i * (k + 1)] += f[i];
fps h = g * f.inv();
for(int i = 1; i <= n; i++){
cout << h[i].val();
if(i == n) cout << "\n";
else cout << " ";
}
return 0;
}
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